Multiple Choice Questions on Quadratic Equations Class 11 OP Malhotra Exe-10H ISC Maths Solutions Ch-10. In this article you would learn to solve mcq types questions on Quadratic Equations. Step by step solutions of latest textbook has been given as latest syllabus. Visit official Website CISCE for detail information about ISC Board Class-11
Quadratic Equation Class 11 OP Malhotra Multiple Choice Questions ISC Math Solutions Ch-10
| Board | ICSE |
| Publications | S Chand |
| Subject | Maths |
| Class | 11th |
| Chapter-10 | Quadratic Equations |
| Writer | OP Malhotra |
| Exe-10(H) | Multiple Choice Questions. |
Multiple Choice Questions on Quadratic Equations
OP Malhotra ISC Class 11 Maths Solutions
Que-1: The geometric mean of the roots of the equation
x2 − 18x + 9 = 0 is
(a) 3 (b) 3√2 (c) 9 (d) 9√2
Sol: (a) 3
Product of roots = c/a = 9
Geometric Mean = √(9) = 3
Que-2: The value of a for which the equation 2x2 + 2√6x + a = 0 has equal roots is
(a) √2 (b) √3 (c) 2 (d) 3
Sol: (d) 3
Equal roots ⇒ Discriminant = 0
b2 − 4ac = 0
(2√6)2 − 4(2)a = 0
24 − 8a = 0 ⇒ a = 3
Que-3: The roots of the quadratic equation x2 − 2√3x − 22 = 0 are
(a) imaginary
(b) real, rational, equal
(c) real, rational, unequal
(d) real, irrational, unequal
Sol: (c) real, rational, unequal
Compare with ax2 + bx + c = 0
a = 1 b = −2√3 c = −22
Discriminant: D = b2 − 4ac
D = (−2√3)2 − 4(1)(−22)
D = 12 + 88 = 100
Since D > 0 and perfect square,
Que-4: The least integer k which makes roots of the equation x2 + 5x + k = 0 imaginary is
(a) 5 (b) 6 (c) 7 (d) 8
Sol: (c) 7
For imaginary roots: D < 0
25 − 4k < 0
k > 6.25
Least integer k = 7
Que-5: If p, q are the roots of the equation x2 + mx + m2 + a = 0, then p2 + pq + q2 + a will be equal to
(a) −m (b) 0 (c) 1 (d) m2 + a
Sol: (b) 0
p + q = −m
pq = m² + a
p² + pq + q² = (p+q)² − pq
= m² − (m² + a) = −a
p² + pq + q² + a = 0
Que-6: If x₁ and x₂ are the real roots of the equation x2 − kx + c = 0, then the distance between the points A(x₁,0) and B(x₂,0) is
(a) √(k2 − c) (b) √(c − k2) (c) √(k2 − 4c) (d) √(k2 + 4c)
Sol: (c) √(k2 − 4c)
Distance = |x₁ − x₂|
= √((x₁ + x₂)² − 4x₁x₂)
= √(k² − 4c)
Que-7: The sum of all the real values of x satisfying the equation 2(x−1)(x + 5x − 50) = 1 is
(a) −5 (b) 14 (c) −4 (d) 16
Sol: (c) -4
x2 + 5x − 50 = (x + 10)(x − 5)
2(x−1)(x + 10)(x − 5) = 1
Checking simple integer values gives the valid real roots whose sum is
Que-8: If α and β are the roots of 4x2 − 2x − 1 = 0, then β =
(a) −1 / 4α (b) −1 / 2α (c) −1 / α (d) 1 / α
Sol: (a) −1 / 4α
αβ = c/a = −1/4
β = (−1/4)/α
β = −1 / 4α
Que-9: Sum of the roots of the equation |x − 3|² + |x − 3| − 2 = 0 is equal to
(a) 2 (b) 4 (c) 6 (d) −2
Sol: (c) 6
Let y = |x − 3|
y² + y − 2 = 0
(y + 2)(y − 1) = 0
y = 1 (since |x−3| ≥ 0)
|x − 3| = 1 ⇒ x = 4, 2
Sum of roots = 4 + 2 = 6
Que-10: If α, β are the roots of x² − px + 1 = 0 and γ is a root of x² + px + 1 = 0, then (α + γ)(β + γ) is
(a) 0 (b) 1 (c) −1 (d) p
Sol: (a) 0
α + β = p , αβ = 1
γ satisfies x² + px + 1 = 0
γ² + pγ +1 =0
(α+γ)(β+γ)
= αβ + γ(α+β) + γ²
= 1 + pγ + γ²
= 1 + (−1) = 0
Que-11: If one root of the equation x² + (3√2 − 2i)x − 6√2i = 0 is 2i, then the other root is
(a) −2i (b) −3/√2 i (c) −3√2 (d) −(3√2 + 2i)
Sol: (c) −3√2
Given root = 2i
Product of roots = c/a
= −6√2 i
2i × other root = −6√2 i
Other root = −3√2
Que-12: If the difference between the roots of the equation x² + ax + 1 = 0 is less than √5, then the possible values of a lie in
(a) (3, ∞) (b) (−∞, −3) (c) (−3, 3) (d) (−3, ∞)
Sol: (c) (−3, 3)
Roots of x² + ax + 1 = 0
Difference of roots
= √(a² − 4)
Given √(a² − 4) < √5
a² − 4 < 5
a² < 9
|a| < 3
−3 < a < 3
Que-13: If the roots of the quadratic equation x² + px + q = 0 are tan 30° and tan 15°, then the value of 2 + q − p is
(a) 1 (b) 2 (c) 3 (d) 0
Sol: (c) 3
tan30° = 1/√3
tan15° = 2 − √3
p = −(sum of roots)
q = product
Calculate ⇒ 2 + q − p = 3
Que-14: If α ≠ β but α² = 5α − 3 and β² = 5β − 3, then the equation whose roots are α/β and β/α is
(a) x² − 5x + 3 = 0
(b) 3x² − 19x + 3 = 0
(c) x² + 5x − 3 = 0
(d) 3x² + 19x − 3 = 0
Sol: (b) 3x² − 19x + 3 = 0
Given α² = 5α − 3
α² − 5α + 3 = 0
Same for β
α + β = 5
αβ = 3
Roots required = α/β , β/α
Sum = (α² + β²)/αβ
= (25 − 6)/3 = 19/3
Product = 1
Equation:
3x² − 19x + 3 = 0
Que-15: If the equations x² + ax + 1 = 0 and x² − x − a = 0 have a real common root b, then the value of b is
(a) 0 (b) 1 (c) −1 (d) 2
Sol: (c) -1
Put x = b in both equations.
b² + ab + 1 = 0
b² − b − a = 0
Subtracting → a(b+1) + (b+1) = 0
(b+1)(a+1) = 0
∴ b = −1
Que-16: If x is a solution of the equation √(2x + 1) − √(2x − 1) = 1 (x > 1/2), then √(4x² − 1) is equal to
(a) 3/4 (b) 1/2 (c) 2√2 (d) 2
Sol: (a) 3/4
Squaring both sides:
2x+1 + 2x−1 − 2√(4x²−1) = 1
4x − 2√(4x²−1) = 1
Solving ⇒ √(4x²−1) = 3/4
Que-17: The quadratic equation 2x² − (a² + 8a − 1)x + a² − 4a = 0 has roots of opposite sign. Then
(a) a ≤ 0 (b) 0 < a < 4 (c) 4 ≤ a < 8 (d) a ≥ 8
Sol: (b) 0 < a < 4
Product of roots = c/a
= (a² − 4a)/2
For opposite sign → product < 0
a(a − 4) < 0
∴ 0 < a < 4
Que-18: If x is a real number, then x / (x² − 5x + 9) lies between
(a) 1/11 and 1
(b) −1 and 1/11
(c) −11 and 1
(d) −1/11 and 1
Sol: (d) −1/11 and 1
x² − 5x + 9 = (x − 2.5)² + 2.75 > 0
By analysis of expression range:
−1/11 < x/(x² − 5x + 9) < 1
Que-19: If one root of a quadratic equation is 1 / (1 + √3), then the quadratic equation is
(a) 2x² + 2x − 1 = 0
(b) 2x² + x − 1 = 0
(c) 2x² − 2x − 1 = 0
(d) 2x² + x + 1 = 0
Sol: (a) 2x² + 2x − 1 = 0
Rationalize root:
x = 1/(1+√3) = (1−√3)/2
Other root = (1+√3)/2
Form equation:
(x − (1−√3)/2)(x − (1+√3)/2) = 0
⇒ 2x² + 2x − 1 = 0
Que-20: If x is real, the minimum value of y = (x² − x + 1)/(x² + x + 1) is
(a) 1/3 (b) 1/2 (c) 2 (d) 3
Sol: (a) 1/3
Using transformation / derivative method, minimum value
= 1/3.
Que-21: The maximum value of 1 / (4x² + 2x + 1) is
(a) 13/4 (b) 5/2 (c) 4/3 (d) 9/5\
Sol: (c) 4/3
Minimum value of denominator gives maximum value of function
⇒ 4/3.
Que-22: If x is real then the value of (x² + 14x + 9) / (x² + 2x + 3) lies in
(a) (−∞, −5) ∪ (4, ∞)
(b) [−5, 4]
(c) (−∞, −4) ∪ (5, ∞)
(d) [−4, 5]
Sol: (b) [−5, 4]
Let y = (x² + 14x + 9)/(x² + 2x + 3)
⇒ y(x² + 2x + 3) = x² + 14x + 9
⇒ (y−1)x² + (2y−14)x + (3y−9) = 0
For real x, discriminant ≥ 0
(2y − 14)² − 4(y − 1)(3y − 9) ≥ 0
⇒ −4y² − 4y + 80 ≥ 0
⇒ (y + 5)(y − 4) ≤ 0
⇒ −5 ≤ y ≤ 4
Que-23: For a ≠ b, if the equations x2 + ax + b = 0 and x2 + bx + a = 0 have a common root, then the value of a + b is
(a) −2 (b) −1 (c) 0 (d) 2
Sol: (b) -1
Let α be the common root.
α² + aα + b = 0 …(1)
α² + bα + a = 0 …(2)
Subtract (1) − (2):
(a − b)α + (b − a) = 0
(a − b)(α − 1) = 0
Since a ≠ b → α = 1
Substitute in (1):
1 + a + b = 0
∴ a + b = −1
Que-24: If the equations k(6x2 + 3) + rx + 2x2 − 1 = 0 and 6k(2x2 + 1) + px + 4x2 − 2 = 0 have both roots common, then the value of (2r − p) is
(a) 0 (b) 1/2 (c) 1 (d) 2
Sol: (a) 0
First equation:
(6k + 2)x² + rx + (3k − 1) = 0
Second equation:
(12k + 4)x² + px + (6k − 2) = 0
Second equation = 2 × First equation
Thus coefficients proportional:
p = 2r
∴ 2r − p = 0
Que-25: If the equation (a + 1)x2 − (a + 2)x + (a + 3) = 0 has roots equal in magnitude but opposite in sign, then the roots of the equation are
(a) ± a (b) ± a/2 (c) ± 3a/2 (d) ± 2a
Sol: (b) ± a/2
If roots are equal in magnitude and opposite in sign, then
sum of roots = 0.
For equation ax2 + bx + c = 0,
Sum of roots = −b/a
Here,
Sum = (a + 2)/(a + 1)
For opposite roots:
(a + 2)/(a + 1) = 0
⇒ a + 2 = 0
⇒ a = −2
Now substitute in equation:
(−1)x2 − 3x + 1 = 0
Solving gives roots:
x = ± a/2
–: End Quadratic Equations Class 11 OP Malhotra Exe-10H ISC Maths Ch-10 Solutions :–
Return to :- OP Malhotra ISC Class-11 S Chand Publication Maths Solutions
Thanks
Please share with your friends
